$f(t) = -t^{3}-2t^{2}-4(g(t))$ $h(x) = 7x^{2}-x-3(g(x))$ $g(x) = -x+2$ $ h(g(-6)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(-6)$ . Then we'll know what to plug into the outer function. $g(-6) = -(-6)+2$ $g(-6) = 8$ Now we know that $g(-6) = 8$ . Let's solve for $h(g(-6))$ , which is $h(8)$ $h(8) = 7(8^{2})-8-3(g(8))$ To solve for the value of $h$ , we need to solve for the value of $g(8)$ $g(8) = -8+2$ $g(8) = -6$ That means $h(8) = 7(8^{2})-8+(-3)(-6)$ $h(8) = 458$